3 ms·
" That's probably sufficient for cases with infrequent read-write cycles, assuming that your floats tend to cover useful time-space coordinates (i.e. where and
by nadam 12y ago
" That's probably sufficient for cases with infrequent read-write cycles, assuming that your floats tend to cover useful time-space coordinates (i.e. where and when a hard drive could plausibly reside). I'm not sure how "space-time coordinates" would work (where is the origin and what is the granularity?), so I don't know if it's likely."
Now I have calculated a little:
The age of the universe is 8 * 10^60 in planck time units:
http://www.wolframalpha.com/input/?i=universe+age+%2F+planck+time+ http://www.wolframalpha.com/input/?i=universe+age+%2F+planck...
So if the simulation is for example something like a cellural automaton with a planck-time (5.4 * 10^(-44) sec) step-size, than assuming we use fix-point numbers, 26 bytes suffice. (25 bytes are just not enough, 256^25 = 1.6 * 10^60)
Suprisingly the size of the universe is 5.4 * 10^61 in planck-length units:
http://www.wolframalpha.com/input/?i=planck+length+universe+size http://www.wolframalpha.com/input/?i=planck+length+universe+...
So again, we need 26 bytes, for a super fine representation. I was one byte off for all components, so we need altogether 104 bytes.
- baddox 12y agoIt sounds like you probably don't need or want floats then. Just a uniform number format for the coordinates should suffice.
- Chinjut 12y agoFor what it's worth, despite the common perception that the Planck length is the the grid-scale on which the world is quantized, or the smallest measurable distance, or various such things, the Planck length is not actually known to have any particular physical significance; it's just the length that falls out of combining various other constants of significance. It's not unreasonable to guess that it will eventually turn out to have some significance in itself, but if there is any, it is as of yet unknown.